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Question Number 149395 by mathdanisur last updated on 05/Aug/21

Answered by Olaf_Thorendsen last updated on 05/Aug/21

∫_a ^b f(x) dx = lim_(n→∞) ((b−a)/n)Σ_(k=1) ^n f(a+k((b−a)/n))  ∫_3 ^5 x^4  dx = lim_(n→∞) ((5−3)/n)Σ_(k=1) ^n (3+k((5−3)/n))^4   ∫_3 ^5 x^4  dx = lim_(n→∞) (2/n)Σ_(k=1) ^n (3+((2k)/n))^4   ⇒ D

$$\int_{{a}} ^{{b}} {f}\left({x}\right)\:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{b}−{a}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({a}+{k}\frac{{b}−{a}}{{n}}\right) \\ $$$$\int_{\mathrm{3}} ^{\mathrm{5}} {x}^{\mathrm{4}} \:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}−\mathrm{3}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}+{k}\frac{\mathrm{5}−\mathrm{3}}{{n}}\right)^{\mathrm{4}} \\ $$$$\int_{\mathrm{3}} ^{\mathrm{5}} {x}^{\mathrm{4}} \:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}+\frac{\mathrm{2}{k}}{{n}}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{D}} \\ $$$$ \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Ser, Thank You

$${Ser},\:{Thank}\:{You} \\ $$

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